The maximum volume flow rate of water is determined as 0.029 m³/s.
<h3>Power of the pump</h3>
The power of the pump is watt is calculated as follows;
1 hp = 745.69 W
7 hp = ?
= 7 x 745.69 W
= 5,219.83 W
<h3>Mass flow rate of water</h3>
η = mgh/P
mgh = ηP
m = ηP/gh
m = (0.82 x 5,219.83)/(9.8 x 15)
m = 29.12 kg/s
<h3>Maximum volume rate</h3>
V = m/ρ
where;
- ρ is density of water = 1000 kg/m³
V = (29.12)/(1000)
V = 0.029 m³/s
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D pad or rb or lb hop this helps
Answer:
power developed by the turbine = 6927.415 kW
Explanation:
given data
pressure = 4 MPa
specific enthalpy h1 = 3015.4 kJ/kg
velocity v1 = 10 m/s
pressure = 0.07 MPa
specific enthalpy h2 = 2431.7 kJ/kg
velocity v2 = 90 m/s
mass flow rate = 11.95 kg/s
solution
we apply here thermodynamic equation that
energy equation that is

put here value with
turbine is insulated so q = 0
so here

solve we get
w = 579700 J/kg = 579.7 kJ/kg
and
W = mass flow rate × w
W = 11.95 × 579.7
W = 6927.415 kW
power developed by the turbine = 6927.415 kW